Two kinds of rocks move along the vibrating table of an electrostatic separator,

Question

Two kinds of rocks move along the vibrating table of an electrostatic separator, with 6 times as many pebbles than rocks. The average pebble has a mass of 4 g and the average rock has a mass of 74 g. Each pebble gains a charge of 8 mC, and the rocks fall 1 m and are between the plates that generate their electric field the whole time.

What is the average charge on a rock?

If the place where the pebbles land is displaced 46 cm horizontally from the spot directly under where they start to fall, how far away from the same spot do the rock land?

What is the magnitude of the electric field between the plates?

Explanation / Answer

given there are 6 times as many pebbles as there are rocks
charge on each pebble, q = 8*10^-3 C
so to maintain net charge neutrality
charge on each rock = Q
Q + 6q = 0
a. Q = -6q = -48*10^-3 C
   so charge on each rock = -48 mC
b. distance of the place where pebbles land from the point where they should have landed vertically = d = 46 cm
   now let assume that the table is at height h
   then time taken for pebbels to fall down = t
   h = 0.5gt^2
   t = sqroot(2h/g)

   now, for a pebble of mass m and charge q
   d = 0.5*at^2
   here a = qE/m
   so d = 0.5*qE*2h/mg = qEh/mg ( where E is electirc field in the region)

   for rocks, Q = 6q
   E, h, g are same
   m = 4g
   M = 74g = 74m/4 = 18.5m

   so distance rocks fall from the vertical spot D = QEh/Mg = 6d/18.5 = 0.324d
   D = 14.918 cm
   sop the rocks will fall at a distance D = 14.918 cm from the vertical spotwhere they should have fallen if they had no charge

c. now,
   h = 1m
   0.46 = qEh/mg = 8*10^-3*E*1/0.004*9.81
   E = 2.2563 V/m ( electric field between the plates)

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